On Friday, 11 November 2016 at 13:39:32 UTC, RazvanN wrote:
It does work, the problem is that [1, 2, 3].sort() is of type SortedRange(int[], "a < b") while r is of type SortedRange(Result, "a < b"). This is a problem if you want to return r in a function which has return type SortedRange(int[], "a < b").
If you absolutely need a `SortedRange(int[], "a < b")` without substitute, then there's no way except building an array that has the numbers sorted, which is always going to be O(n) unless you can place the two input arrays adjacent into memory beforehand somehow. However, if you just want the pre-chain ranges and post-chain ranges be a compatible type, you can use a class wrapper, such as RandomAccessFinite - thus, the assumeSorted result will be `SortedRange(RandomAccessFinite, "a < b")`. Note, though, that even though the algorithmic complexity will be O(1), every access to the wrapped range will go through a virtual method call, so it will affect performance in practice. There is no way around this because the type pulls in the underlying range's behavior, and e.g. SortedRange!(int[]) knows that the underlying elements are arranged linearly in memory, SortedRange of a chain knows that it first has to check which sub-range any operation will refer to, and SortedRange of a RandomAccessFinite knows that it just has to pass the request to a virtual class method which hides the underlying implementation.
